# Chemistry

Calculate the [h30+] in a 0.010M solution of Sr(OH)2

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1. 0.010M in Sr(OH)2.
OH^- is twice that.
(H3O^+)(OH^-) = Kw = 1E-14
Solve for (H3O^+)

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2. 5*10•14

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3. because Sr(OH)2, do 0.010 x 2

ans = 0.020

pOH = -log(0.020)
pOH = 1.7
[finding pH because Sr(OH)2]
pH = 14 - 1.7
pH = 12.30

10^-12.30 = 5.0x10^-13

hope this helps

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